Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2009, Article ID 535209, 17 pages
doi:10.1155/2009/535209
Research Article
Singular Positone and Semipositone
Boundary Value Problems of Nonlinear Fractional
Differential Equations
Chengjun Yuan,1, 2 Daqing Jiang,1 and Xiaojie Xu1
1
2
School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, Jilin, China
School of Mathematics and Computer, Harbin University, Harbin 150086, Heilongjiang, China
Correspondence should be addressed to Chengjun Yuan, ycj7102@163.com
Received 8 January 2009; Accepted 15 April 2009
Recommended by Victoria Vampa
We present some new existence results for singular positone and semipositone nonlinear fractional
boundary value problem Dα0 ut μatft, ut, 0 < t < 1, u0 u1 u 0 u 1 0, where
μ > 0, a, and f are continuous, α ∈ 3, 4 is a real number, and Dα0 is Riemann-Liouville fractional
derivative. Throughout our nonlinearity may be singular in its dependent variable. Two examples
are also given to illustrate the main results.
Copyright q 2009 Chengjun Yuan et al. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
Fractional calculus has played a significant role in engineering, science, economy, and other
fields. Many papers and books on fractional calculus, and fractional differential equations
have appeared recently, see 1–9. It should be noted that most of papers and books on
fractional calculus, are devoted to the solvability of initial fractional differential equations
see 3, 4. Here, we consider positive solutions of nonlinear fractional differential equation
conjugate boundary value problem involving Riemann-Liouville derivative:
Dα0 ut μatft, ut,
0 < t < 1,
u0 u1 u 0 u 1 0,
1.1
where μ > 0, a, and f are continuous. α ∈ 3, 4 is a real number, and Dα0 is the RiemannLiouville fractional derivative.
2
Mathematical Problems in Engineering
It is well known that in mechanics the boundary value problem 1.1 where α 4
describes the deflection of an elastic beam rigidly fixed at both ends. The integer order
boundary value problem 1.2 has been studied extensively. For details, see for instance, the
papers 10–13 and the references therein. In 10, 12, Yao considered
u t λft, ut,
0 < t < 1,
u0 u1 u 0 u 1 0,
1.2
and using a Krasnosel’skii fixed-point theorem, derived a λ-interval such that, for any λ lying
in this interval, the beam equation has existence and multiplicity on positive solution. In this
paper, we will consider a more general situation, namely, the boundary value problem 1.1.
To the best of our knowledge, there have been few papers which deal with the boundary
value problem 1.1 for nonlinear fractional differential equation.
In this paper, in analogy with boundary value problems for differential equations
of integer order, we firstly derive the corresponding Green’s function named the fractional
Green’ function. Consequently problem 1.1 is reduced to an equivalent Fredholm integral
equation of the second kind. Finally, using Krasnosel’skii’s fixed-point theorems, the
existence of positive solutions are obtained.
2. Preliminaries
For completeness, in this section, we will demonstrate and study the definitions and some
fundamental facts of Riemann-Liouville derivatives of fractional order which can be found
in 5.
Definition 2.1 see 5, Definition 2.1. The integral
α
fx I0
1
Γα
x
0
ft
x − t1−α
dt,
x > 0,
2.1
where α > 0, is called the Riemann-Liouville fractional integral of order α.
Definition 2.2 see 5, page 36-37. For a function fx given in the interval 0, ∞, the
expression
Dα0 fx
n x
ft
d
1
dt,
α−n1
Γn − α dx
0 x − t
2.2
where n α 1, α denotes the integer part of number α, is called the Riemann-Liouville
fractional derivative of order α.
From the definition of Riemann-Liouville derivative, for μ > −1, we have
Dα0 xμ
Γ 1μ
xμ−α ,
Γ 1μ−α
2.3
giving in particular Dα0 xα−m 0, m 1, 2, 3, . . . , N, where N is the smallest integer greater
than or equal to α.
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3
Lemma 2.3. Let α > 0, then the differential equation
Dα0 ut 0
2.4
has solutions ut c1 tα−1 c2 tα−2 · · · cn tα−n , ci ∈ R, i 1, , 2 . . . , n, as unique solutions, where
n is the smallest integer greater than or equal to α.
α
u u, from Lemma 2.3, we deduce the following statement.
As Dα0 I0
Lemma 2.4. Let α > 0, then
α
I0
Dα0 ut ut c1 tα−1 c2 tα−2 · · · cn tα−n ,
2.5
for some ci ∈ R, i 1, 2, . . . , n, n is the smallest integer greater than or equal to α.
The following Krasnosel’skii’s fixed-point theorem will play a major role in our next analysis.
Theorem 2.5 see 6. Let X be a Banach space, and let P ⊂ X be a cone in X. Assume that Ω1 , Ω2
are open subsets of X with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 , and let S : P → P be a completely continuous operator
such that, either
1 Sw ≤ w, w ∈ P ∩ ∂Ω1 , Sw ≥ w, w ∈ P ∩ ∂Ω2 , or
2 Sw ≥ w, w ∈ P ∩ ∂Ω1 , Sw ≤ w, w ∈ P ∩ ∂Ω2 .
Then S has a fixed-point in P ∩ Ω2 \ Ω1 .
3. Green’s Function and Its Properties
In this section, we derive the corresponding Green’s function for boundary-value problem
1.1, and obtain some properties of Green’s function.
Lemma 3.1. Let ht ∈ C0, 1 be a given function, then the boundary-value problem,
Dα0 ut ht,
0 < t < 1, 3 < α ≤ 4,
u0 u1 u 0 u 1 0,
3.1
has a unique solution
ut 1
Gt, shsds,
3.2
0
where
⎧
α−2
α−1
α−2
1 ⎨t 1 − s s − t α − 21 − ts t − s , 0 ≤ s ≤ t ≤ 1,
Gt, s Γα ⎩tα−2 1 − sα−2 s − t α − 21 − ts,
0 ≤ t ≤ s ≤ 1.
Here Gt, s is called Green’s function of boundary-value problem 3.1.
3.3
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Mathematical Problems in Engineering
Proof. By means of the Lemma 2.4, we can reduce 3.1 to an equivalent integral equation
ut c1 tα−1 c2 tα−2 c3 tα−3 c4 tα−4 t
t − sα−1
hsds.
Γα
0
3.4
From u0 u1 u 0 u 1 0, we have c3 c4 0 and
c1 c2 1
1 − sα−2 2s − αs − 1
hsds,
Γα
0
3.5
1
α − 11 − s
Γα
0
α−2
s
hsds.
Therefore, the unique solution of 3.1 is
1
tα−1 1−sα−2 2s−αs−1
hsds ut Γα
0
1
Γα
t 1
1
tα−2 α−11−sα−2 s
hsds Γα
0
t
t−sα−1
hsds
0 Γα
tα−1 1−sα−2 2s−αs−1 tα−2 α−11−sα−2 s t−sα−1 hsds
0
tα−1 1−sα−2 2s−αs−1 tα−2 α−11−sα−2 s hsds
t
1
Gt, shsds.
0
3.6
The proof is finished.
Lemma 3.2. The function Gt, s defined by 3.3 has the following properties:
1 Gt, s G1 − s, 1 − t, for
t, s ∈ 0, 1;
2 tα−2 1 − t2 qs ≤ Gt, s ≤ α − 1qs and Gt, s ≤ α − 1α − 2/Γαtα−2 1 − t2
for t, s ∈ 0, 1,
where qs α − 2/Γαs2 1 − sα−2 .
Proof. Observing the expression of Gt, s, it is clear that Gt, s G1 − s, 1 − t for t, s ∈ 0, 1.
In the following, we consider ΓαGt, s.
Mathematical Problems in Engineering
5
For 0 ≤ s ≤ t ≤ 1, we have
ΓαGt, s t − sα−1 − t − tsα−2 t − s α − 21 − tt − tsα−2 s
t − s t − sα−2 − t − tsα−2 α − 21 − tt − tsα−2 s
−t − sα − 2
t−ts
xα−3 dx α − 2t − tsα−2 1 − ts
t−s
≥ −t − sα − 2t − tsα−3 t − ts − t − s α − 2t − tsα−2 1 − ts
3.7
−t − sα − 2t − tsα−3 1 − ts α − 2t − tsα−2 1 − ts
α − 2t − tsα−3 1 − ts−t − s t − ts
≥ α − 2t − tsα−2 1 − t2 s2
α − 2tα−2 1 − t2 s2 1 − sα−2 ,
and
ΓαGt, s −t − sα − 2
t−ts
xα−3 dx α − 2t − tsα−2 1 − ts
t−s
≤ −t − sα − 2t − sα−3 t − ts − t − s α − 2t − tsα−2 1 − ts
−t − sα − 2t − sα−3 1 − ts α − 2t − tsα−2 1 − ts
α − 21 − ts t − tsα−2 − t − sα−2
≤ α − 22 1 − t2 s2 t − tsα−3
≤ α − 1α − 2tα−3 1 − t2 s2 1 − sα−3
≤ α − 1α − 2s2 1 − sα−2 .
For 0 ≤ t ≤ s ≤ 1, since α > 3, we have
ΓαGt, s t − tsα−2 s − t α − 21 − ts
≥ α − 2t − tsα−2 1 − ts
α − 2tα−2 1 − sα−2 1 − ts
≥ α − 2tα−2 1 − t2 s2 1 − sα−2 ,
3.8
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Mathematical Problems in Engineering
ΓαGt, s tα−2 1 − sα−2 s − t α − 21 − ts
≤ tα−2 1 − sα−2 s α − 2s
≤ α − 1t1 − sα−2 s
≤ α − 1s2 1 − sα−2
≤ α − 2α − 1s2 1 − sα−2 .
3.9
Thus tα−2 1 − t2 qs ≤ Gt, s ≤ α − 1qs, for t, s ∈ 0, 1. Combining Gt, s G1 − s, 1 − t,
we have
Gt, s ≤ α − 1q1 − t α − 1α − 2 α−2
t 1 − t2 , for t, s ∈ 0, 1.
Γα
3.10
This completes the proof.
We note that ut is a solution of 1.1 if and only if
ut μ
1
Gt, sasfusds,
0 ≤ t ≤ 1.
3.11
0
For our constructions, we will consider the Banach space E C0, 1 equipped with
standard norm u max0≤t≤1 ut, u ∈ X. We define a cone K by
tα−2 1 − t2
u ∈ X | ut ≥
u, t ∈ 0, 1, α ∈ 3, 4 .
α−1
K
3.12
Define an integral operator A : K → X by
Aut μ
1
Gt, sasfusds,
0 ≤ t ≤ 1, u ∈ K.
3.13
0
Notice from 3.13 and Lemma 3.2 that, for u ∈ K, Aut ≥ 0 on 0, 1 and
Aut μ
1
Gt, sasfs, usds ≤ μ
0
then Au ≤
1
0
α − 1qsasfs, usds .
1
0
α − 1qsasfs, usds,
3.14
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7
On the other hand, we have
Aut μ
1
Gt, sasfs, usds
0
≥μ
1
tα−2 1 − t2 qsasfs, usds
0
tα−2 1 − t2
μ
≥
α−1
≥
1
3.15
α − 1qsasfs, usds
0
tα−2 1 − t2
Au.
α−1
Thus, AK ⊂ K. In addition, standard arguments show that A is completely continuous.
4. Singular Positone Problems
In this section we present some new result for the singular problem
Dα0 ut μatft, ut,
0 < t < 1, 3 < α ≤ 4,
u0 u1 u 0 u 1 0,
4.1
where μ > 0 and nonlinearity f may be singular at u 0.
Using Theorem 2.5 we establish the following main result.
Theorem 4.1. Suppose that the following conditions are satisfied.
a ∈ C0, 1 ∩ L1 0, 1 with a > 0 on 0, 1
f : 0, 1 × 0, ∞ −→ 0, ∞ is continuous,
⎧
⎪
⎪
⎪ft, u ≤ gu hu on 0, 1 × 0, ∞ with g > 0
⎪
⎪
⎨
continuous and nonincreasing on 0, ∞, h ≥ 0
⎪
⎪
⎪
h
⎪
⎪
non decreasing on 0, ∞,
⎩continuous on 0, ∞ and
g
∃K0 with g xy ≤ K0 gxg y
∀x > 0, y > 0,
1
a0 μα − 1 qsasg sα−2 1 − s2 ds < ∞,
4.2
4.3
4.4
4.5
4.6
0
r
1
2
> K0 a0 g
∃r > 0 with
,
gr hr
α−1
4.7
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Mathematical Problems in Engineering
⎧
1 ⎪
⎪
choose and fix it and a continuous,
T here exists 0 < θ <
⎪
⎪
2
⎪
⎪
⎪
⎪
⎨nonincreasing function g : 0, ∞ −→ 0, ∞, and a continuous
1
⎪
h1
⎪
⎪
nondecreasing on 0, ∞
function h1 : 0, ∞ −→ 0, ∞ with
⎪
⎪
⎪
g1
⎪
⎪
⎩
and with ft, u ≥ g1 u h1 u for t, u ∈ θ, 1 − θ × 0, ∞,
⎧
⎪
⎪
⎨∃ 0 < R1 < r < R2 with i 1, 2,
1−θ
Ri g1 θα /α − 1Ri
⎪
⎪
<
μ
Gσ, sasds,
⎩
g1 Ri g1 θα /α − 1Ri g1 Ri h1 θα /α − 1Ri θ
4.8
4.9
here Gt, s is Green’s function and
1−θ
Gσ, sasds sup
θ
1−θ
Gt, sasds.
4.10
t∈0,1 θ
Then 4.1 has two nonnegative solutions ui with R1 < u1 < r < u2 < R2 and ui t > 0 for
t ∈ 0, 1, i 1, 2.
Proof. First we will show that there exists a solution u2 to 4.1 with u2 t > 0 for t ∈ 0, 1 and
r < u2 < R2 . Let
Ω1 {u ∈ E : u < r},
Ω2 {u ∈ E : u < R2 }.
4.11
We now show
Au < u
for K ∩ ∂Ω1 .
4.12
To see this, let u ∈ K ∩ ∂Ω1 . Then u ||u||0,1 r and ut ≥ tα−2 1 − t2 /α − 1r for
t ∈ 0, 1. So for t ∈ 0, 1, we have
Aut μ
1
Gt, sasfs, usds
0
≤μ
1
α − 1qsas gus hus ds
0
1
hus
ds
μ α − 1qsasgus 1 gus
0
Mathematical Problems in Engineering
≤μ
1
9
α − 1qsasg
0
sα−2 1 − s2
r
α−1
hr
ds
1
gr
1
hr
r
1
μ α − 1qsasg sα−2 1 − s2 ds
α−1
gr
0
1
a0 K02 g
gr hr .
α−1
≤ K0 g
4.13
This together with 4.7 yields
Au Au0,1 < r u,
4.14
for K ∩ ∂Ω2 .
4.15
so 4.12 is satisfied.
Next we show
Au > u
To see this, let u ∈ K ∩ ∂Ω2 so u u0,1 R2 , and let ut ≥ tα−2 1 − t2 /α − 1R2 for
t ∈ 0, 1.
We have
Auσ μ
1
Gσ, sasfs, usds
0
≥μ
1−θ
Gσ, sas g1 us h1 us ds
θ
1−θ
h1 us
ds
Gσ, sasg1 us 1 μ
g1 us
θ
⎫
⎧
1−θ
⎨
h1 sα−2 1 − s2 /α − 1 R2 ⎬
Gσ, sas 1 ≥ g1 R2 μ
ds
⎩
θ
g1 sα−2 1 − s2 /α − 1 R2 ⎭
≥ g1 R2 μ
1−θ
θ
4.16
h1 θα /α − 1R2 ds.
Gσ, sas 1 g1 θα /α − 1R2 This together with 4.9 yields
Auσ > R2 u.
4.17
Thus ||Au|| > ||u||, so 4.15 is held.
Now Theorem 2.5 implies that A has a fixed-point u2 ∈ K ∩ Ω2 \ Ω1 , that is, r ≤
u2 u2 0,1 ≤ R and u2 t ≥ qtr for t ∈ 0, 1. It follows from 4.12 and 4.15 that
r, u2 /
R2 , so we have r < ||u2 || < R2 .
u2 /
10
Mathematical Problems in Engineering
Similarly, if we put
Ω1 {u ∈ E : u < R1 },
Ω2 {u ∈ E : u < r},
4.18
we can show that there exists a solution u1 to 4.1 with u1 t > 0 for t ∈ 0, 1 and R1 < ||u1 ||
< r.
This completes the proof of Theorem 4.1.
Similar to the proof of Theorem 4.1, we have the following result.
Theorem 4.2. Suppose that 4.2–4.8 hold. In addition suppose
⎧
⎪
⎪
⎨∃0 < R1 < r with
1−θ
R1 g1 θα /α − 1R1
⎪
⎪
<μ
Gσ, sqsds.
⎩
g1 R1 g1 θα /α − 1R1 g1 R1 h1 θα /α − 1R1
θ
4.19
Then 4.1 has a nonnegative solution u1 with R1 < u1 < r and u1 t > 0 for t ∈ 0, 1.
Remark 4.3. If in 4.19 we have R1 > r, then 4.1 a nonnegative solution u2 with r < u2 < R2
and u2 t > 0 for t ∈ 0, 1.
It is easy to use Theorem 4.2 and Remark 4.3 to write theorems which guarantee the
existence of more than two solutions to 4.1. We state one such result.
Theorem 4.4. Suppose that 4.2–4.6 and 4.8 hold. Assume that ∃m ∈ {1, 2, . . .} and constants
Ri , ri i 1, . . . , m, with r1 > b0 , and
0 < R1 < r1 < R2 < r2 < · · · < Rm < rm .
4.20
In addition suppose for each i 1, . . . , m that
ri
1
> K02 a0 g
,
gri hri α−1
4.21
and
Ri g1 θα /α − 1Ri
<μ
g1 Ri g1 θα /α − 1Ri g1 Ri h1 θα /α − 1Ri
1−θ
Gσ, sasds
4.22
θ
hold. Then 4.1 has nonnegative solutions y1 , . . . , ym with yi t > 0 for t ∈ 0, 1.
Example 4.5. Consider the boundary value problem
Dα0 ut σ u−a t ub t ,
t ∈ 0, 1, 3 < α ≤ 4,
u0 u1 u 0 u 1 0,
0 < a < 1 < b,
4.23
Mathematical Problems in Engineering
11
where σ ∈ 0, σ0 is such that
σ0 ≤
1
,
2a1
4.24
here
a1 α − 1α − 2
Γα
1
1
−α
sα−2 1 − s2 sα−2 1 − s2
ds sα−21−a 1 − s21−a ds < ∞.
0
0
4.25
Then 4.23 has two solutions u1 , u2 with u1 t > 0, u2 t > 0 for t ∈ 0, 1, i 1, 2.
To see this we will apply Theorem 4.1 with here 0 < R1 < 1 < R2 will be chosen below
gu g1 u u−a ,
hu h1 u ub ,
at σ,
K0 1, θ 1
.
4
4.26
Clearly 4.2–4.6 and 4.8 hold, and a0 σa1 /α − 1a . Now 4.7 holds with r 1 since
1
r
1
≥ a1 σ0 > K02 a0 α − 1a K02 a0 g
.
gr hr 2
α−1
4.27
Finally notice 4.9 is satisfied for R1 small and R2 large since
R1a
Ri
i
−→ 0,
g1 Ri 1 h1 θα /α − 1Ri / g1 θα /α − 1Ri 1 α − 1−ab θαab Rab
i
4.28
as R1 → 0, R2 → ∞, since b > 1. Thus all the conditions of Theorem 4.1 are satisfied so
existence is guaranteed.
5. Singular Semipositone Problems
In this section we present a new result for the singular semipositone problem:
Dα0 ut μatft, ut,
0 < t < 1, 3 < α ≤ 4,
u0 u1 u 0 u 1 0,
5.1
where μ > 0 and nonlinearity f may be singular at u 0.
Before we prove our main result, we first state a result.
Lemma 5.1. Suppose a ∈ L1 0, 1 with a > 0 on 0, 1. Then the boundary value problem,
Dα0 ut atet,
0 < t < 1, 3 < α ≤ 4,
u0 u1 u 0 u 1 0,
5.2
12
Mathematical Problems in Engineering
has a solution w with
wt ≤
tα−2 1 − t2
C0
α−1
for t ∈ 0, 1,
5.3
here
C0 α − 12 α − 2
Γα
1
asesds.
5.4
0
In fact, from Lemma 3.1, 5.2 has solution
wt 1
Gt, sasesds.
5.5
tα−2 1 − t2
α − 1α − 2tα−2 1 − t2
asesds C0 .
Γα
α−1
0
5.6
0
According to Lemma 3.2, we have
wt ≤
1
The above Lemma together with Theorem 2.5 establish our main result.
Theorem 5.2. Suppose that the following conditions are satisfied.
q ∈ C0, 1 ∩ L1 0, 1
with q > on 0, 1.
⎧
⎨f : 0, 1 × 0, ∞ −→ R is continuous and there exists
⎩a function e ∈ C0, 1, 0, ∞
with ft, u et ≥
5.7
for t, u ∈ 0, 1 × 0, ∞,
5.8
⎧
⎪
f ∗ t, u ft, u et ≤ gu hu on 0, 1 × 0, ∞ with g > 0
⎪
⎪
⎪
⎪
⎨
continuous and nonincreasing on 0, ∞, h ≥ 0
⎪
⎪
⎪
h
⎪
⎪
nondecreasing on 0, ∞,
⎩continuous on 0, ∞ and
g
∃K0 with g xy ≤ K0 gxg y
∀x > 0, y > 0,
1
a0 Gσ, sqsg sα−2 1 − s2 ds < ∞,
5.9
5.10
5.11
0
∃r > μC0 with
g
r
≥ μK0 a0 ,
r − μC0 /α − 1 1 hr/gr
5.12
Mathematical Problems in Engineering
⎧
1 ⎪
⎪
choose and fix it and a continuous,
T here exists 0 < θ <
⎪
⎪
2
⎪
⎪
⎪
⎪
⎨nonincreasing function g1 : 0, ∞ −→ 0, ∞, and a continuous
⎪
h1
⎪
⎪
nondecreasing on 0, ∞
function h1 : 0, ∞ −→ 0, ∞ with
⎪
⎪
⎪
g1
⎪
⎪
⎩
and with ft, u et ≥ g1 u h1 u for t, u ∈ θ, 1 − θ × 0, ∞,
13
5.13
and ∃R > r with
Rg1 θα /α − 1R
≤μ
g1 Rg1 θα /α − 1R g1 Rh1 θα /α − 1R
1−θ
Gσ, sqsds,
5.14
θ
here > 0 is any constant (choose and fix it) so that 1 − μC0 /R ≥ (note exists since R > r > μC0
in fact we can have 1 − μC0 /r) ) and Gt, s is Green’s function and
1−θ
Gσ, sasds sup
θ
1−θ
Gt, sasds.
5.15
t∈0,1 θ
Then 5.1 has a solution y ∈ C0, 1 ∩ C2 0, 1 with yt > 0 for t ∈ 0, 1.
Proof. To show that 5.1 has a nonnegative solution we will look at the boundary value
problem
Dα0 yt μqtf ∗ t, yt − φt ,
0 < t < 1, 3 < α ≤ 4,
y0 y1 y 0 y 1 0,
5.16
where
φt μMwt,
0 ≤ t ≤ 1,
5.17
w is as in Lemma 5.1.
We will show, using Theorem 2.5 , that there exists a solution y1 to 5.16 with y1 t >
φt for t ∈ 0, 1. If this is true then ut y1 t − φt, 0 ≤ t ≤ 1 is a nonnegative solution
positive on 0, 1 of 5.1, since
Dα0 ut Dα0 y1 t − Dα0 φt −μqtf ∗ t, y1 t − φt μqtet
−μqt f t, y1 t − φt et μqtet
−μqtf t, y1 t − φt
−μqtft, ut,
5.18
0 < t < 1.
As a result, we will concentrate our study on 5.16. Let E, K as in Section 2, and let
Ω1 {u ∈ E : u < r},
Ω2 {u ∈ E : u < R}.
5.19
14
Mathematical Problems in Engineering
Next let A : K ∩ Ω2 \ Ω1 → E be defined by
Ay t μ
1
Gt, sqsf ∗ s, ys − φs ds,
0 ≤ t ≤ 1.
5.20
0
In addition, standard argument shows that AP ⊂ P and A is completely continuous.
We now show
Ay ≤ y
for K ∩ ∂Ω1 .
5.21
To see this, let y ∈ K ∩ ∂Ω1 . Then y y0,1 r and yt ≥ tα−2 1 − t2 /α − 1r for
t ∈ 0, 1. Now for t ∈ 0, 1, the Lemma 5.1 implies
tα−2 1 − t2
tα−2 1 − t2
tα−2 1 − t2 r−μ
C0 ≥
r − μC0 > 0,
α−1
α−1
α−1
yt − φt ≥
5.22
so for t ∈ 0, 1 we have
Ay t μ
1
Gt, sasf ∗ s, ys − φs ds
0
≤μ
1
qsas g ys − φs h ys − φs ds
0
h ys − φs
μ qsasg ys − φs 1 ds
g ys − φs
0
1
sα−2 1 − s2 hr
≤ μ qsasg
ds
1
r − μC0
α−1
gr
0
1
5.23
1
r − μC0
hr
≤ μK0 g
qsasg sα−2 1 − s2 ds
1
α−1
gr
0
r − μC0
h{r}
μK0 a0 g
.
1
α−1
g{r}
This together with 5.12 yields
Ay Ay
≤ r y,
0,1
5.24
so 5.21 is satisfied.
Next we show
Ay ≥ y
for K ∩ ∂Ω2 .
5.25
Mathematical Problems in Engineering
15
To see this let y ∈ K ∩ ∂Ω2 so y y0,1 R and yt ≥ tα−2 1 − t2 /α − 1R for t ∈ 0, 1.
Also for t ∈ 0, 1 we have
tα−2 1−t2
tα−2 1−t2
R−μC0
α−1
α−1
μC0
tα−2 1−t2
tα−2 1−t2
R 1−
≥
R.
≥
α−1
R
α−1
yt−φt yt−μwt ≥
5.26
As a result
yt − φt ≥
θα
R
α−1
for t ∈ θ, 1 − θ.
5.27
We have
Ay σ μ
1
Gσ, sqsf ∗ s, ys − φs ds
0
≥μ
1−θ
Gσ, sqs g1 ys − φs h1 ys − φs ds
0
μ
1−θ
h1 ys − φs
1 ds
g1 ys − φs
Gσ, sqsg1 ys − φs
0
≥ μg1 R
1−θ
0
5.28
h1 θα /α − 1R
ds.
Gσ, sqs 1 g1 θα /α − 1R
This together with 5.14 yields
Ay σ ≥ R y.
5.29
Thus Ay ≥ y, so 5.25 is held.
Now Theorem 2.5 implies that A has a fixed-point y ∈ K ∩ Ω2 \ Ω1 , that is, r ≤
y y0.1 ≤ R and yt ≥ tα−2 1 − t2 r for t ∈ 0, 1. Thus yt is a solution of 5.16 with
yt > φt for t ∈ 0, 1. Thus 5.1 has a positive solution ut yt−φt > for t ∈ 0, 1.
Example 5.3. Consider the boundary value problem
Dα0 yt μ y−a t yb t − 1 ,
y0 y1 y 0 y 1 0,
0 < t < 1, 3 < α ≤ 4,
5.30
0 < a < 1 < b,
where μ ∈ 0, μ0 is such that
1/α
α − 12 α − 2μ0
α − 1 2μ0 a0
≤ 1,
2Γα
5.31
16
Mathematical Problems in Engineering
here
a0 1
1
qsasg sα−2 1 − s2 ds s1−aα−1 1 − sα−1 ds < ∞.
0
5.32
0
Then 5.30 has a solution y with yt > 0 for t ∈ 0, 1.
To see this we will apply Theorem 5.2 with here R > 1 will be chosen later, in fact here
we choose R > 1 so that 1/2 works, i.e., we choose R so that 1 − μ/2ΓαR ≥ 1/2,
g y g1 y y−a ,
aK0 1, 1
,
2
θ
1
,
4
h y h1 y yb ,
C0 α − 12 α − 2
Γα
at 1, et t−1/2 ,
1
s−1/2 ds 0
α − 22 α − 2
.
2Γα
5.33
Clearly 5.7–5.11 and 5.13 hold. Now 5.12 holds with r 1 since
μK0 a0 μa0 < μ0 a0 ≤
1 1−μ0 C0 a 1 1−μc0 a
≤
2
α−1
2 α−1
r
,
1 hr/gr g r −μMC0 /α−1
5.34
from 5.31. Finally notice 5.14 is satisfied for R large since
Rg1 θα /α − 1R
R1a
−→ 0,
g1 Rg1 θα /α − 1R g1 Rh1 θα /α − 1R 1 α − 1−ab θα ab Rab
5.35
as R → ∞, since b > 1. Thus all the conditions of Theorem 5.2 are satisfied so existence is
guaranteed.
Acknowledgments
This paper is supported by Key Subject of Chinese Ministry of Education no. 109051 and
Scientific Research Fund of Heilongjiang Provincial Education Department no. 11544032.
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